SOLUTION: Three boys have marbles in the ratio 19 : 5 : 3. If the boy with the least number has 9 marbles, how many marbles does the boy with the greatest number have?
(A) 57 (B) 62
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(A) 57 (B) 62
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Question 252881: Three boys have marbles in the ratio 19 : 5 : 3. If the boy with the least number has 9 marbles, how many marbles does the boy with the greatest number have?
(A) 57 (B) 62 (C) 48 (D) 54 (E) 70
You can put this solution on YOUR website! let A be the one with the largest number.
let B be the one with the middle number.
let C be the one with the smallest number.
let a = the number of marbles that A has.
let b = the number of marbles that B has.
let c = the number of marbles that C has.
you get:
a:b:c = 19:5:3
C has 9 marbles.
9/3 = 3 which means that C has 3 times the marbles as it's ratio number.
to keep the ratios intact, all others in the chain have to be multiplied by 3.
you get:
a:b:c = 19:5:3 = 57:15:9
this means that A has 57 marbles while B has 15 marbles while C has 9 marbles.
you can also use the ratio formula to find the correct value.
b:c = 5:3
This is the same as b/c = 5/3 since a fraction is a ratio of 2 numbers.
you have b/c = 5/3.
you know that c = 9.
the equation becomes:
b/9 = 5/3
multiply both sides of this equation by 9 to get:
b = (9*5)/3 = 45/3 = 15.
now that you know b, you can solve for a in the same manner.
a:b = 19:5
this is the same as a/b = 19/5.
you have a/b = 19/5.
you know that b is 15.
the equation becomes:
a/15 = 19/5
multiply both sides of this equation by 15 to get:
a = (15*19)/5 = 285/5 = 57
you wind up with:
a = 57
b = 15
c = 9
your question was how many marbles does the boy with the greatest number have?
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